Show that
cot^2(p/7) + cot^2(2p/7) + cot^2(3p/7) = 5.
See this: http://www.mathisfunforum.com/viewto...d=79371#p79371
It was proved that $\displaystyle \tan^2{\frac{\pi}{7}},\tan^2{\frac{2\pi}{7}},\tan^ 2{\frac{3\pi}{7}}$ are the roots of the equation $\displaystyle x^3-21x^2+35x-7=0$.
Hence $\displaystyle \cot^2{\frac{\pi}{7}},\cot^2{\frac{2\pi}{7}},\cot^ 2{\frac{3\pi}{7}}$ are the roots of the equation $\displaystyle \left(\frac{1}{x}\right)^3-21\left(\frac{1}{x}\right)^2+35\left(\frac{1}{x}\r ight)-7=0$, i.e. of the equation $\displaystyle 7x^3-35x^2+21x-1=0$.